How much `PCl_(5)` must be added to a one little vessel at `250^(@)C` in order to obtain a concentration of 0.1 mole of `Cl_(2)` at equilibrium. `K_(c)` for `PCl_5(g)hArrPCl_(3)(g)+Cl_(2)(g)` is `0.0414`

Options

A. 3.415 mole
B. 34.15 mole
C. 0.03415 mole
D. 0.3415 mole

Answer

D

Explanation

`PCl_(5) Leftrightarrow PCl_(3)+CI_(2)`
`{:(,"initial",a,0,0),(,eq:, (a-0.1),0.1,0.1):}`
`K_(c)=((0.1)/(1) xx (0.1))/(1))/((a-0.1)/(1)),` a=0.3415 mole